# Solve complex equationcomplex equation z^4+9=0 solve in z

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You may write the equation as difference of squares such that:

`z^4 - (3i)^2 = 0` (use complex number theory to write `i^2=-1` )

Use the special product (a-b)(a+b) to write difference of squares `a^2-b^2` such that:

`(z^2 - 3i)(z^2 + 3i)=0`

`z^2 - 3i = 0 =gt z^2 = 3i`

Use the standard form of complex number such that `z = x + iy =gt z^2 = x^2 + 2ixy - y^2`

Hence, `x^2 + 2ixy - y^2 = 3i`

You need to equate real parts such that:

`x^2 - y^2 = 0 =gt x^2 = y^2 =gt x = +-y`

You need to equate imaginary parts such that:

`2xy = 3`

If `x = y =gt 2x^2 = 3 =gt x^2 = 3/2 =gt x_(1,2) = +-sqrt6/2 =gt y_(1,2) = +-sqrt6/2`

`x = -y =gt -2x^2 = 3 =gt x^2 = -3/2 =gt x_(1,2) = +-(i*sqrt6)/2 =gt y_(1,2) = +-(i*sqrt6)/2`

`` **Hence, the solutions to the equation are: `z_(1,2) = +-sqrt6/2 +- i*(+-sqrt6/2), z_(3,4) = -+sqrt6/2 +- i*sqrt6/2` .**